=> Magnetic moment of Vanadium is 3.9 BM . Find out oxidation state of vanadium ?
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Magnetic moments are often used in conjunction with electronic spectra to gain information about the oxidation number and stereochemistry of the central metal ion in coordination complexes.
A common laboratory procedure for the determination of the magnetic moment for a complex is the Gouy method which involves weighing a sample of the complex in the presence and absence of a magnetic field and observing the difference in weight. A template is provided for the calculations involved.
For first row transition metal ions in the free ion state, i.e. isolated ions in a vacuum, all 5 of the 3d orbitals are degenerate.
A simple crystal field theory approach to the bonding in these ions assumes that when they form octahedral complexes, the energy of the d orbitals are no longer degenerate but are split such that two orbitals, the d x2-y2 and the d z2 (e g subset) are at higher energy than the d xy , d xz , d yz orbitals (the t2g subset).
For octahedral ions with between 4 and 7 d electrons, this gives rise to 2 possible arrangements called either high spin/weak field or low spin/strong field respectively. The energy gap is dependent on the position of the coordinated ligands in the SPECTROCHEMICAL SERIES.
See an interactive JAVA applet for examples.
Note: For CHEM1902, we assume that all Co(III), d 6 complexes are octahedral and LOW spin, i.e. t 2g 6 .
In tetrahedral complexes, the energy levels of the orbitals are again split, such that the energy of two orbitals, the d x2-y2 and the d z2 (e subset) are now at lower energy (more favoured) than the remaining three d xy , d xz , d yz (the t2 subset) which are destabilised.
Tetrahedral complexes are ALL high spin since the difference between the 2 subsets of energies of the orbitals is much smaller than is found in octahedral complexes.
The usual relationship quoted between them is: Δ tet ≈ 4/9 Δ oct .
Square planar complexes are less commmon than tetrahedral and for CHEM1902 we will assume that the only ions forming square planar complexes are d 8 e.g. Ni(II), Pd(II), Pt(II), etc. d 8 can therefore be either square planar or tetrahedral. As with octahedral complexes, the energy gap between the dxy and dx2 -y 2 is Δoct and these d 8 systems are all considered strong field / low spin complexes hence they are all diamagnetic, μ=0 Bohr Magneton (B.M.).
The formula used to calculate the spin-only magnetic moment can be written in two forms; the first based on the number of unpaired electrons, n, and the second based on the total electron spin quantum number, S. Since for each unpaired electron, n=1 and S=1/2 then the two formulae are clearly related and the answer obtained must be identical.
μ so = √n(n+2) B.M.
μ so = √4S(S+1) B.M. - a variation of this will be introduced in the
second year Inorganic course: μ S+L = √{4S(S+1) + L(L+1)} B.M.
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A common laboratory procedure for the determination of the magnetic moment for a complex is the Gouy method which involves weighing a sample of the complex in the presence and absence of a magnetic field and observing the difference in weight. A template is provided for the calculations involved.
For first row transition metal ions in the free ion state, i.e. isolated ions in a vacuum, all 5 of the 3d orbitals are degenerate.
A simple crystal field theory approach to the bonding in these ions assumes that when they form octahedral complexes, the energy of the d orbitals are no longer degenerate but are split such that two orbitals, the d x2-y2 and the d z2 (e g subset) are at higher energy than the d xy , d xz , d yz orbitals (the t2g subset).
For octahedral ions with between 4 and 7 d electrons, this gives rise to 2 possible arrangements called either high spin/weak field or low spin/strong field respectively. The energy gap is dependent on the position of the coordinated ligands in the SPECTROCHEMICAL SERIES.
See an interactive JAVA applet for examples.
Note: For CHEM1902, we assume that all Co(III), d 6 complexes are octahedral and LOW spin, i.e. t 2g 6 .
In tetrahedral complexes, the energy levels of the orbitals are again split, such that the energy of two orbitals, the d x2-y2 and the d z2 (e subset) are now at lower energy (more favoured) than the remaining three d xy , d xz , d yz (the t2 subset) which are destabilised.
Tetrahedral complexes are ALL high spin since the difference between the 2 subsets of energies of the orbitals is much smaller than is found in octahedral complexes.
The usual relationship quoted between them is: Δ tet ≈ 4/9 Δ oct .
Square planar complexes are less commmon than tetrahedral and for CHEM1902 we will assume that the only ions forming square planar complexes are d 8 e.g. Ni(II), Pd(II), Pt(II), etc. d 8 can therefore be either square planar or tetrahedral. As with octahedral complexes, the energy gap between the dxy and dx2 -y 2 is Δoct and these d 8 systems are all considered strong field / low spin complexes hence they are all diamagnetic, μ=0 Bohr Magneton (B.M.).
The formula used to calculate the spin-only magnetic moment can be written in two forms; the first based on the number of unpaired electrons, n, and the second based on the total electron spin quantum number, S. Since for each unpaired electron, n=1 and S=1/2 then the two formulae are clearly related and the answer obtained must be identical.
μ so = √n(n+2) B.M.
μ so = √4S(S+1) B.M. - a variation of this will be introduced in the
second year Inorganic course: μ S+L = √{4S(S+1) + L(L+1)} B.M.
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Shaina11:
But what is final answer ? I mean oxidation state ?
The oxidation state, sometimes referred to as oxidation number, is an indicator of the degree of oxidation (loss of electrons) of an atom in a chemical compound.
no I asked numerical answer Ji
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Answer:
+3.
Explanation:
BM is 3.9 the main integar is 3 that means the upaired number of electrons is 3 so number of unpaired electrons are 3 that means 3 are involved in the bonding which give oxidation number that is 3 so +3 is answer.
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